Question

$\frac{E^2}{{\mu }_0}$ has the dimensions of (where $E$ is electric field and ${\mu }_0$ is permeability of free space)

• A. $\left[M{L}^3{T}^{-2}\right]$
• B. $\left[M^{-1}{L}^{2}T{A}^{-2}\right]$
• C. $[ML{T}^{-4}]$
• D. $\left[M^2{L}^3{T}^{-2}{A}^2\right]$

Answer 1

Answer: (C)

$\frac{E^2}{{\mu }_0}=\frac{{\epsilon }_0{E}^2}{{\epsilon }_0{\mu }_0}$
where ${\epsilon }_0$ is the permittivity of free space
$\therefore$ Energy density (i.e. energy per unit volume)
of an electric field is $u_E=\frac{1}{2}{\epsilon }_0{E}^2$
Speed of light in vacuum, $c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$
$\therefore \frac{E^2}{{\mu }_0}=\frac{2u_E}{\frac{1}{c^2}}$
$=2u_E{c}^2$
$\left[\frac{E^2}{{\mu }_0}\right]=\left[M{L}^{-1}{T}^{-2}\right]{\left[L{T}^{-1}\right]}^2$
$=\left[ML{T}^{-4}\right]$